Let's tackle this step-by-step.

Given Information

• Albert rolls a six-sided die $n = 30$ times.
• He observes a 6 occurring $x = 6$ times out of 30 rolls.
• Null Hypothesis $H_0: p = 0.1667$, where $p$ is the true probability of rolling a 6 with a fair die.
• Alternative Hypothesis $H_a: p > 0.1667$ (Albert believes he rolls a 6 more often than expected).

The test is a one-sided (right-tailed) test.

Step (b): Calculate the p-value

To find the p-value, we can use the binomial test or approximate with a normal distribution, given that $n$ is sufficiently large.

1. Expected Probability $p_0 = 0.1667$.
2. Observed Proportion $\hat{p} = \frac{x}{n} = \frac{6}{30} = 0.2$.
3. Standard Error of $\hat{p}$ under the null hypothesis: $\text{SE} = \sqrt{\frac{p_0 (1 - p_0)}{n}} = \sqrt{\frac{0.1667 \times 0.8333}{30}}$

Let's compute this standard error and use it to find the z-score and the p-value.The p-value for this test is approximately $0.312$.

Step (c): Conclusion at the 5% Significance Level

Since the p-value (0.312) is much larger than the significance level of 0.05, we do not have enough evidence to reject the null hypothesis. Therefore, this sample does not provide sufficient evidence at the 5 percent level to conclude that Albert rolls a 6 more often than expected by chance.

Would you like a more detailed explanation of any part of this process?

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Here are some related questions that can expand your understanding:

1. How would the result change if the sample size $n$ were larger?
2. What if Albert rolled a 6 more than 6 times out of 30—how would that affect the p-value?
3. Why do we use a normal approximation in this test?
4. How is a one-tailed test different from a two-tailed test in hypothesis testing?
5. What other methods can we use to verify Albert's claim?

Tip: In hypothesis testing, a larger sample size generally provides more reliable evidence for or against a hypothesis.